You have found the following ages (in years) of 6 turtles. The turtles are randomly selected from the 32 turtles at your local zoo: $ 20,\enspace 32,\enspace 22,\enspace 86,\enspace 68,\enspace 28$ Based on your sample, what is the average age of the turtles? What is the variance? You may round your answers to the nearest tenth.
Solution: Because we only have data for a small sample of the 32 turtles, we are only able to estimate the population mean and variance by finding the sample mean $({\overline{x}})$ and sample variance $({s^2})$ To find the sample mean , add up the values of all $6$ samples and divide by $6$ $ {\overline{x}} = \dfrac{\sum\limits_{i=1}^{{n}} x_i}{{n}} = \dfrac{\sum\limits_{i=1}^{{6}} x_i}{{6$ To compensate for this underestimation, rather than simply averaging the squared deviations from the mean , we total them and divide by $n - 1$ $ {s^2} = \dfrac{\sum\limits_{i=1}^{{n}} (x_i - {\overline{x}})^2}{{n - 1}} $ $ {s^2} = \dfrac{{515.29} + {114.49} + {428.49} + {1874.89} + {640.09} + {216.09}} {{6 - 1}} $ $ {s^2} = \dfrac{{3789.34}}{{5}} = {757.87\text{ years}^2} $ We can estimate that the average turtle at the zoo is 42.7 years old. There is a variance of 757.87 years $^2$.